Question

Write an equation of the sine function with the following:
amp=7
period=3(pi)
phase shift=(pi)
vertical shift=-7

Answer 1

If period is \[3\pi\] then wouldn't k= 2(3π)/k

??

Answer 2

wait

2(3π)/k = π

Answer 3

Given\[y =a \sin k(x -p)+q\]a is the amplitude, k is used to determine the period and is given by\[k =\frac{ 2\pi }{ period }\], pis the phase shift, and q is the equation of the axis of the curve (the mean value).

Answer 4

@workin_daily your k is incorrect. Look at my note.

Answer 5

@workin_daily do you understand?

Answer 6

Is your phase shift of π to the left or to the right. You didn't indicate.

Answer 7

I'm looking at my book and it says the equation will be y=Asin(kθ+c)+h

Answer 8

also mentions that K: 2π/k = π so if that was the case then k would equal 2

Answer 9

So I thought

2(3π)/k = π

Answer 10

\[y =A \sin [k(\theta -c)+h\]Exactly the same thing. There are 26 letters in the Latin alphabet. Which letter you choose to represent what, is irrelevant. c is the same as p and h is the same as q. Same thing!!!!!

Answer 11

No, \[k =\frac{ 2\pi }{ period }\]as I said before.

Answer 12

I wasn't reffering to the q or h, it was the k you had outside of the parentheses, which I would assume would result in a different answer than if it was inside, such as my book shows.

Answer 13

The k always has to be factored before identifying the phase shift.

Answer 14

but if we put k \[k(\theta+c)=k \theta +k c\]
this will mean we are no longer having a phase shift of c but ok kc,wich is not what is asked.

HENCE WE HAVE TO REPRESENT AS
\[k \theta+c\]

Answer 15

@Jonask if it's (kθ + c) then the phase shift is c/k.

Answer 16

If your period is 3π, then your\[k =\frac{ 2\pi }{ 3\pi }=\frac{ 2 }{ 3 }\]See?

Answer 17

Okay, so what about the book's formula for it?

Answer 18

Yes \[period =\frac{ 2\pi }{ k }\]but if you rearrange this equation for k, then\[k =\frac{ 2\pi }{ period }\]

Answer 19

is there any difference between shifting \[c\]units to the right and a phase shift of c units to the right

Answer 20

I never said your book is wrong all I was trying to tell you is that if you solve for k by then k = 2pi/period. Do you understand now?

Answer 21

@Jonask yes it does matter which is why I asked @workin_daily if it is a phase shift of π to the left or to the right but I never received a reply..

Answer 22

All it gave me was

phase shift = π

Answer 23

if \[y=\sin(x)\]
shift \[\pi\] units to the left
then\[y=\sin(x+\pi)\]

Answer 24

Then @workin_daily since it's positive pi, they mean to the right.

Answer 25

\[\sin(2x)\]
shifted pi units right\[\sin(2x+\pi)\]

Answer 26

No @Jonask if the 2 is inside the parentheses, with π to the right it should be\[\sin (2x - 2\pi )=\sin 2(x -\pi )\]

Answer 27

you said phase shift should be c/k if the formula being used was (kθ + c) right?

Answer 28

So then, y=7sin(2/3+(c/.667))+h

the .667 being 2/3

Answer 29

i strongly believe that in \[\sin(k \theta)\] if \[\phi\]is a phase shift then\[\sin(k \theta+\phi)\]
is the new function as the book said

Answer 30

\[y=Asin(k θ+c)+h\]

Answer 31

Nvm, what I said doesn't really make much sense

Answer 32

If we have (kθ + c) = k(θ + c/k) then to find the phase shift what one really does is either sets kθ + c = 0 or θ + c/k = 0 and solve for θ. Either way θ = -c/k which is c/k to the left because of the negative sign. I never said your book was wrong, I 'm simply trying to demonstrate that there are different ways of expressing the same thing. Understand?

Answer 33

I always teach my students different approaches to solving the same problem. It's not just about the different methods (although often one method is easier or more elegant then the other), but more importantly it is understanding the key concepts of a problem.

Answer 34

I get that, by book slightly differs from what my notes say.
In y=Asin(kθ+c)+h
Does h stand for the vertical shift? Because in my notes I have
y=Asin(kθ+h)+V
with the V standing for my vertical shift
(I'm pretty sure, on this, but I'm pretty tired so just double checking)

Answer 35

oh yeah, nvm, just read it in my book too haha. Guess I'm just getting myself confused

Answer 36

So then tell me your final answer so that I can rest assured that you now know what to do.

Answer 37

so my phase shift is p or pi/1 which should = c/k
k=2/3 so it's
c/(2/3)=(pi)/1

Answer 38

c=(pi)/(2/3)

Answer 39

Exactly! See you didn't believe when I tried to tell you so many times. No faith! LOL

Answer 40

So you still haven't shown me the final equation.

Answer 41

y = 7sin( 2/3 + π/.667)-7

Answer 42

Not perfect! Please try again and this time also without any decimals. Although that is not why it's wrong.

Answer 43

y = 7sin((2/3)θ + π/(2/3))-7

Answer 44

I should have said that is not the only reason. That phase shift is still incorrect because of two reasons. Do you know what they are?

Answer 45

Everything else is right. It's just the c value that is wrong.

Answer 46

You said I was correct when I put "c=(pi)/(2/3)"

Answer 47

Oh I didn't see that you had pi/2, I thought I saw pi times 2. So then it's my fault for telling you that it was right.

Answer 48

\[c =\frac{ 2\pi }{ 3 }\]
\[c \neq \frac{ \frac{ \pi }{ 2 } }{ 3 }\]

Answer 49

Think about it. If you try to solve for θ with what you have you won't get a phase shift of π to the right. Will you?

Answer 50

Okay, this is what I put before
so my phase shift is p or pi/1 which should = c/k
k=2/3 so it's
c/(2/3)=(pi)/1

Answer 51

**edit** (pi) not p

Answer 52

\[y =7\sin (\frac{ 2 }{ 3 }\theta -\frac{ 2\pi }{ 3 })-7\]OR\[y =7\sin \frac{ 2 }{ 3 }(\theta -\pi)-7\]Now if you solve for theta, you will get a phase shift of positive pi, which is what you were given.

Kapish?!

Answer 53

I see it, was just having a really long retard moment haha. I don't know why c was giving me so much trouble, now that I stand back and look at it, it's very easy.

Answer 54

(2/3) * (π/1) = (2π/3)

Answer 55

Not at all! When you're learning concepts for the first time, it's OK experience difficulty. That's why I'm always so patient with my students when teaching new lessons. @workin_daily you did just fine!

Answer 56

Question, where did the "-" come from? Shouldn't it be
y=7sin(2/3 θ + 2π/3 )−7

Answer 57

No, because if the phase shift is to the positive (right) then when you solve for theta, you should get c/k. But if the phase shift is negative (left), then when you solve for theta, you should get -c/k. In the question you asked, phase shift was positive pi, so then when you solve for theta you should get positive pi. If you have what you're suggesting, then solving for theta would yield negative pi, and that was not what we were given.

Answer 58

Understood?!

Answer 59

I think so

Answer 60

Practice makes perfect so do more questions and eventually you will understand.

Answer 61

Well I have plenty more to do, doesn't help that I was out sick for two days of class.. -_-

Answer 62

Well, whenever you're not in school for valid reasons, it is your responsibility to catch up. You can borrow notes from friends or classmates, also find out the homework from them.

Answer 63

If you need anything else, let me know. There is someone who has been waiting for my help so I will help them now and then I have to log out. But if you need anything else, message me and I will be sure to help you the next time I log in.

Answer 64

This problem was actually from the class I missed. Sadly I wasn't ableto get notes before this weekend due to a range of reason so I have been trying to learn the missed material the best I can. And thanks for the help, I'm sure I will have plenty more questions later on.

Answer 65

Actually she doesn't need my help right now so if there isn't anything else right now, I will sign out.

Answer 66

I'm good for now, 2am and class in a few hours, need to get at least some sleep.. thanks again

Answer 67

Definitely to get sleep! Like I said message me if you need anything later. Goodnight!